Dual Nature of Radiation

As we know,

$\begin{aligned} F & =q v B=m \frac{v^2}{R} \Rightarrow v=\frac{q}{m} B R . \\ \mathrm{K} . \mathrm{E} . & =\frac{1}{2} m v^2=\frac{1}{2} \frac{e^2 B^2 R^2}{m}=74.4 \mathrm{~keV} . \end{aligned}$

Energy of a photon $=\frac{h c}{\lambda}=49.5 \mathrm{~keV}$.

Binding energy $=74.4-49.5 \mathrm{~keV}=24.9 \mathrm{~keV}$

To determine the velocity of the electron emitted from the gun, we need to apply the concept of energy conservation. Specifically, the kinetic energy (KE) gained by the electron will be equal to the electrostatic potential energy (PE) provided by the potential difference (V) between the filament and the plate.

The electrostatic potential energy can be expressed as:

$ PE = eV $

where:

• $ e $ is the charge of the electron, approximately $ 1.6 \times 10^{-19} \text{ C} $
• $ V $ is the potential difference, which is given as $ 4000 \text{ V} $

The kinetic energy of the electron can be expressed as:

$ KE = \frac{1}{2} mv^2 $

where:

• $ m $ is the mass of the electron, approximately $ 9.1 \times 10^{-31} \text{ kg} $
• $ v $ is the velocity of the electron we need to determine.

Equating the kinetic energy to the potential energy, we have:

$ eV = \frac{1}{2} mv^2 $

Rearranging the equation to solve for the velocity $ v $ gives:

$ v = \sqrt{\frac{2eV}{m}} $

Substituting the given values:

$ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 4000}{9.1 \times 10^{-31}}} $

$ v = \sqrt{\frac{1.28 \times 10^{-15}}{9.1 \times 10^{-31}}} $

$ v = \sqrt{1.41 \times 10^{15}} $

$ v \approx 3.75 \times 10^7 \text{ m/s} $

Therefore, the correct option is:

Option D: $ 3.75 \times 10^7 \mathrm{~m} / \mathrm{s} $

To determine the rate of change of the de-Broglie wavelength of the electron at time $t$, we need to start with the expression for the de-Broglie wavelength, which is given by

$\lambda = \frac{h}{p}$

where $h$ is Planck's constant and $p$ is the momentum of the electron.

Since the electron experiences a constant electric field of $2E$, it undergoes a uniform acceleration. The force $ F $ on the electron is given by

$F = e \cdot 2E = 2eE$

This force causes the electron to accelerate, so the acceleration $a$ is given by

$a = \frac{F}{m} = \frac{2eE}{m}$

Initially, the electron is at rest, so its velocity $v$ at time $t$ can be obtained using the equation of motion:

$v = u + at = 0 + \left(\frac{2eE}{m}\right) t = \frac{2eE t}{m}$

The momentum $p$ of the electron at time $t$ is given by

$p = mv = m \left( \frac{2eE t}{m} \right) = 2eE t$

Now substituting the expression for momentum into the de-Broglie wavelength equation, we get

$\lambda = \frac{h}{p} = \frac{h}{2eE t}$

To find the rate of change of de-Broglie wavelength with respect to time, we take the derivative of $\lambda$ with respect to $t$:

$\frac{d\lambda}{dt} = \frac{d}{dt} \left( \frac{h}{2eE t} \right) = h \cdot \frac{d}{dt} \left( \frac{1}{2eE t} \right) = h \cdot \left( \frac{-1}{(2eE t)^2} \cdot 2eE \right) = \frac{-h}{2eE} \cdot \frac{1}{t^2} \cdot 2eE = \frac{-h}{t^2}$

Thus, the rate of change of de-Broglie wavelength of the electron at time $t$ is

$\frac{d\lambda}{dt} = \frac{-h}{2eE t^2}$

So the correct answer is:

Option A: $-\frac{h}{\left(2eEt^2\right)}$

$e{V_0} = h(\nu - {\nu _0}) = h\nu - h{\nu _0}$

${V_0} = \left( {{h \over e}} \right)\nu - \left( {{h \over e}} \right){\nu _0}$ ..... (i)

From Eq. (i), it follows that the slope of the graph is ${h \over e}$.

Therefore,

h = e $\times$ slope = 1.6 $\times$ 10^$-$19 $\times$ ${{16.5 - 0} \over {(8 - 4) \times {{10}^{15}}}}$

$ = {{1.6 \times {{10}^{ - 19}} \times 16.5} \over {4 \times {{10}^{ - 5}}}} = 6.6 \times {10^{ - 34}}$ J-s.

The de-Broglie wavelength, $\lambda = {h \over {mv}}$

Here, ${\lambda _e} = {h \over {{m_e}{e \over 3}}}$ and ${\lambda _p} = {h \over {{m_p}c}}$

Given, ${\lambda _e} = {\lambda _p}$

So, ${h \over {{m_e}{c \over 3}}} = {h \over {{m_p}c}}$

$ \Rightarrow {{{m_e}} \over {{m_p}}} = 3$

Ratio of $KE = {{{K_e}} \over {{K_p}}} = {{{1 \over 2}{m_e}v_e^2} \over {{1 \over 2}{m_p}v_p^2}} = {1 \over 3}$

($\because$ ${{{m_e}} \over {{m_p}}} = 3,\,{v_e} = {c \over 3}$ and ${v_p} = c$)

Minimum wavelength for X-ray is given by

${\lambda _{\min }} = {{12375} \over V}\mathop A\limits^o $

As the accelerating voltage V is decreased, $\lambda$_min increases. The intensity of emitted radiation is determined by the number of electrons bombarding the target.

$\therefore$ Accelerating voltage does not change the intensity of X-rays emitted.

Kinetic energy in photoelectric effect can be given as

KE = W $-$ W₀ [W $\to$ incident energy, W₀ $\to$ work function]

Given, (KE)₁ = 1.5 $-$ 0.5 = 1 eV

(KE)₂ = 2.5 $-$ 0.5 = 2.0 eV

$\because$ ${{{{(KE)}_1}} \over {{{(KE)}_2}}} = {1 \over {2}}$

$ \Rightarrow {{{1 \over 2}mv_1^2} \over {{1 \over 2}mv_2^2}} = {1 \over 2} \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{1 \over 2}} $

$\therefore$ ${v_1}:{v_2} = 1 :\sqrt 2 $

So, the ratio of maximum speeds of emitted electrons is $1 :\sqrt 2 $.